At a large university, the mean amount spent by students for cell phone service is $38.90 per month with a standard deviation of $3.64 per month. consider a group of 44 randomly chosen university students. what is the probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1?

There is a 7.18% probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

At a large university, the mean amount spent by students for cell phone service is $38.90 per month with a standard deviation of $3.64 per month.

This means that

Consider a group of 44 randomly chosen university students.  What is the probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1?

By the Central Limit Theorem, this means that now we have to use the standard deviation of the mean, since we are working with a sampling distribution of the sample mean. So:

This probability is:

--------

This probability is 1 subtracted by the pvalue of Z when

has a pvalue of 0.9641. This means that .

--------

This probability is the pvalue of Z when .

has a pvalue of 0.0359. So .

So

There is a 7.18% probability that the mean amount of their monthly cell phone bills differs from the mean for the university by more than $1.

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